ECE600 F13 rv Functions of random variable mhossain - Rhea

Random Variables and Signals

Topic 8: Functions of Random Variables

We often do not work with the random variable we observe directly, but with some function of that random variable. So, instead of working with a random variable X, we might instead have some random variable Y=g(X) for some function g:R → R.
In this case, we might model Y directly to bet f $$ _Y $$ (y), especially if we do not know g. Or we might have a model for X and find f $$ _Y $$ (y) (or p $$ _Y $$ (y)) as a function of f $$ _X $$ (or p $$ _X $$ and g.
We will discuss the latter approach here.

More formally, let X be a random variable on (S,F,P) and consider a mapping g:R → R. Then let Y $(\omega)=$ g(X( $\omega))$ ∀ $\omega$ ∈ S.
We normally write this as Y=g(X).

Graphically,

Fig 1: Mapping from S to X to Y under g

Is Y a random variable? We must have Y $^{-1}$ (A) ≡ { $\omega$ ∈ S: Y $(\omega)$ ∈ A} = { $\omega$ ∈ S: g(X $(\omega)$ ) ∈ A} be an element of F ∀A ∈ B(R) (Y must be Borel measurable).
We will only consider functions g in this class for which Y $^{-1}$ (A) ∈ F ∀A ∈ B(R), so that if Y=g(X) for some random variable X, Y will be a random variable.

What is the distribution of Y? Consider 3 cases:

X discrete, Y discrete
X continuous, Y discrete
X continuous, Y continuous

Note: you cannot have a continuous Y from a discrete X.

Case 1: X and Y Discrete

Let $$ R_X $$ ≡ X(S) be the range space of X and math>R_Y</math>≡ g(X(S)) be the range space of Y. Then the pmf of Y is

p

$ _Y $

(y) = P(Y=y) = P(g(X)=y)

But this means that

p_Y(y) = \sum_{x\in\mathcal{R}_X:g(x)=y}p_X(x)\;\;\forall y\in\mathcal{R}_Y

Example $\quad$ Let X be the value rolled on a die and

Y = \begin{cases} 1 & \mbox{if}\;X\;\mbox{is odd} \\ 0 & \mbox{if}\;X\;\mbox{is even} \end{cases}

Then R $$ _X $$ = {0,1,2,3,4,5,6} and R $$ _Y $$ = {0,1} and g(x) = x % 2.

Now

p_Y(y) = \sum_{x\in\mathcal{R}_X:g(x)=y}p_X(x)

\begin{align} \Rightarrow p_Y(0) &= p_X(2)+p_X(4)+p_X(6) \\ p_Y(1) &= p_X(1)+p_X(3)+p_X(5) \end{align}

Case 2: X Continuous, Y Discrete

The pmf of Y in this case is

p

$ _Y $

(y) = P(g(X)=y) = P(X ∈ D

$ _Y $

) =
where D

$ _Y $

≡ {x ∈ R: g(x)=y} ∀y ∈ R

$ _Y $

Then,

p_Y(y) = \int_{D_y}f)X(x)dx

Example Let g(x) = u(x - x $$ _0 $$ ) for some x $$ _0 $$ ∈ R, and let Y=g(X). Then $$ R_Y $$ = {0,1} and

D

$ _0 $

= {x ∈ R: x < x

$ _0 $

} = (-∞, x

$ _0 $

) D

$ _1 $

= {x ∈ R: x ≥ x

$ _0 $

} = [ x

$ _0 $

, ∞)

So,

p_Y(y) = \begin{cases} \int_{-\infty}^{x_0} f_X(x)dx & y=0\\ \\ \int_{x_0}^{-\infty} f_X(x)dx & y=1 \end{cases}

Case 3: X and Y Continuous

We will discuss 2 methods for finding f $$ _Y $$ in this case.

Approach 1
First, find the cdf F $$ _Y $$ .

F

$ _Y $

(y) = P(g(X) ≤ y) = P(X ∈ D

$ _y $

)
where D

$ _y $

= {x ∈ R: g(x) ≤ y}.

Then

F_Y(f) = \int_{D_y}f_X(x)dx

Differentiate F $$ _Y $$ to get f $$ _y $$ .

You can find D $$ _Y $$ graphically or analytically

Example

Fig 2: This plot of g(x) can be used to derive D

$ _Y $

graphically

For y = y $$ _1 $$ and y = y $$ _2 $$ ,

\begin{align} D_{y_1} &= \{x:\;x \leq x_1\} \\ D_{y_2} &= \{x:\;x\leq x_2'\} \cup \{x:\;x_2''<x\leq x_2'''\} \end{align}

Then

\begin{align} F_Y(y_1) &= \int_{-\infty}^{x_1}f_X(x)dx \\ \\ F_Y(y_2) &= \int_{-\infty}^{x_2'}f_X(x)dx + \int_{x_2''}^{x_2'''}f_X(x)dx \end{align}

Example Y = aX + b, a,b ∈ R, a ≠ 0

F

$ _Y $

(y) = P(aX + b ≤ y)

So,

\begin{align} D_y&=\{x\in\mathbb R: x\leq\frac{y-b}{a}\}\quad\mbox{if}\;a>0 \\ D_y&=\{x\in\mathbb R: x\geq\frac{y-b}{a}\}\quad\mbox{if}\;a<0 \end{align}

Then

F_Y(y)=\begin{cases} \int_{-\infty}^{\frac{y-b}{a}}f_X(x)dx & \mbox{if }\;a>0 \\ \\ \int_{\frac{y-b}{a}}^{-\infty}f_X(x)dx & \mbox{if }\;a<0 \end{cases}

Example Y = X $$ ^2 $$

Fig 3: Y = X

$ ^2 $

For y < 0, D $$ _y $$ = ø
For y ≥ 0,

\begin{align} F_Y(y) &= P(X^2\leq y) \\ &= P(-\sqrt{y} <X\leq \sqrt{y}) \end{align}

So,

D_y = (-\sqrt{y},\sqrt{y})

and

F_Y(y) = \int_{-\sqrt{y}}^{\sqrt{y}}f_X(x)dx

Approach 2

Use a formula for f $$ _y $$ in terms of f $$ _X $$ . To derive the formula, assume the inverse function g $^{-1}$ exists, so if y = g(x), then x = g $^{-1}$ (y). Also assume g and g $^{-1}$ are differentiable. Then, if Y = g(X), we have that

f_Y(y) = \frac{f_X(g^{-1}(y))}{|\frac{dy}{dx}|_{x=g^{-1}(y)}}

Proof:
First consider g monotone (strictly monotone) increasing

Fig 4: Function g is strictly increasing on its domain.

Since {y < Y ≤ y + Δy} = {x < X ≤ x + Δx}, we have that P(y < Y ≤ y + Δy) = P(x < X ≤ x + Δx).

Use the following approximations:

P(y < Y ≤ y + Δy) ≈ f $$ _Y $$ (y)Δy
P(x < X ≤ x + Δx) ≈ f $$ _X $$ (x)Δx

Fig 5: P(y < Y ≤ y + Δy) ≈ f

$ _Y $

(y)Δy

Since the left hand sides are equal,

f_Y(y)\Delta y \approx f_X(x)\Delta x

Now as Δy → 0, we also have that Δx → 0 since g is continuous, and the approximations above become equalities. We rename Δy, Δx as dy and dx respectively, so letting Δy → 0, we get

\begin{align} f_Y(y)dy &= f_X(x)dx \\ \Rightarrow f_Y(y)&=f_X(x)\frac{dx}{dy} \end{align}

We normally write this as

f_Y(y) = \frac{f_X(g^{-1}(y))}{\frac{dy}{dx}|_{x=g^{-1}(y)}}

A similar derivation for g monotone decreasing gives us the general result for invertible g:

f_Y(y) = \frac{f_X(g^{-1}(y))}{|\frac{dy}{dx}|_{x=g^{-1}(y)}}

Note this result can be extended to the case where y = g(x) has n solutions x $$ _1 $$ ,...,x $$ _n $$ , in which case,

f(Y(y) = \sum_{i=1}^n\frac{f_X(x_n)}{|\frac{dy}{dx}|_{x=x_n}}

For example, if Y = X $$ ^2 $$ ,

x_1 = -\sqrt{y},\;\;x_2 = \sqrt{y}

\Rightarrow f_Y(y) = \frac{f_X(-\sqrt{y})}{2\sqrt{y}}+\frac{f_X(\sqrt{y})}{2\sqrt{y}}

References

M. Comer. ECE 600. Class Lecture. Random Variables and Signals. Faculty of Electrical Engineering, Purdue University. Fall 2013.

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ECE600 F13 rv Functions of random variable mhossain - Rhea

Contents

Case 1: X and Y Discrete

Case 2: X Continuous, Y Discrete

Case 3: X and Y Continuous

References

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